在 Xcode 中使用 NSURLConnection 发送 JSON 数据
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Send JSON data with NSURLConnection in Xcode
提问by Nick
I am trying to send JSON data to a web server via the code below. For some reason, the request does not seem to be going out. What does it look like I am missing? Also the result from NSURLConnection (retStr) is always empty?
我正在尝试通过以下代码将 JSON 数据发送到 Web 服务器。出于某种原因，该请求似乎没有发出。看起来我失踪了什么？NSURLConnection (retStr) 的结果也总是空的？
NSDictionary *data = [NSDictionary dictionaryWithObject:@"test sending ios" forKey:@"value1"];
NSError *error;
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:data options:kNilOptions error:&error];
NSURL *url = [NSURL URLWithString:@"http://webserveraddress"];
NSMutableURLRequest *req = [NSMutableURLRequest requestWithURL:url cachePolicy:nil timeoutInterval:60];
[req setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[req setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[req setValue:[NSString stringWithFormat:@"%d", [jsonData length]] forHTTPHeaderField:@"Content-Length"];
[req setHTTPMethod:@"POST"];
[req setHTTPBody:jsonData];
NSString *retStr = [[NSString alloc] initWithData:[NSURLConnection sendSynchronousRequest:req returningResponse:nil error:nil] encoding:NSUTF8StringEncoding];
采纳答案by Nicolas Manzini
To send simple data in post vars to your webserver running php you simply do this in
要将 post vars 中的简单数据发送到运行 php 的网络服务器，您只需在
Example
例子
NSString * key = [NSString stringWithFormat:@"var1=%@&var2=%@&var3=%@",@"var1String" ,@"var2string" ,[NSnumber numberWithBool:YES]];
NSURL * url = [NSURL URLWithString:@"http://webserver.com/yourScriptPHP.php"];
NSMutableURLRequest * request = [NSMutableURLRequest requestWithURL:url];
[request setHTTPMethod:@"POST"];
[request setHTTPBody:[key dataUsingEncoding:NSUTF8StringEncoding]];
[[NSURLConnection alloc] initWithRequest:request delegate:self];
// this is for you to be able to get your server answer.
// you will need to make your class a delegate of NSURLConnectionDelegate and NSURLConnectionDataDelegate
myClassPointerData = [[NSMutableData data] retain];
Implement
实施
-(void)connection:(NSURLConnection *)connection didReceiveData:(NSData *)data {
[myClassPointerData appendData:data]
}
-(void)connection:(NSURLConnection *)connection DidFinishLoading {
// do what you want with myClassPointerData the data that your server did send you back here
// for info on your server php script you just need to do: echo json_encode(array('var1'=> $var1, 'var2'=>$var2...));
// to get your server sending an answer
}
回答by codeburn
Instead of sending as JSON, you could simply send it the normal way - by setting the request's HTTPBody with POST method.
您可以简单地以正常方式发送，而不是作为 JSON 发送 - 通过使用 POST 方法设置请求的 HTTPBody。
Only differenc you need to make if you need to make a call such as 'http://abc.example.com/user_profile.json?user[first_name]=fdffdf&user[last_name]=dffdf'
如果您需要拨打诸如“http://abc.example.com/user_profile.json?user[first_name]=fdffdf&user[last_name]=dffdf”之类的电话，则只需要进行区分
is, you need to have your dictionary keys have a prefix. That is, 'first_name=fdffdf' needs to be changed to 'user[first_name]=fdffdf'.
是，你需要让你的字典键有一个前缀。即'first_name=fdffdf'需要改为'user[first_name]=fdffdf'。
Try this bit of code to change your parameter dictionary to the required format for JSON..
试试这段代码，将您的参数字典更改为 JSON 所需的格式。
for (id key in [self allKeys]) {
NSString *newKey = [NSString stringWithFormat:@"%@[%@]", parent, key];
[result setObject:[self objectForKey:key] forKey:newKey];
}
回答by ganesh manoj
once check like this
一旦像这样检查
ASIFormDataRequest *request=[ASIFormDataRequest requestWithURL:[NSURL URLWithString:@"ur url"]];
ASIFormDataRequest *request=[ASIFormDataRequest requestWithURL:[NSURL URLWithString:@"ur url"]];
[request setPostValue:appdelegate.userid forKey:@"userid"];
[request setPostValue:self.nameLbl.text forKey:@"username"];
[request setPostValue:self.website.text forKey:@"website"];
NSLog(@"~~~~~~ request~~~~~ %@",request);
[request setTimeOutSeconds:5000];
[request startSynchronous];
